Lec%2025_Full

# Lec%2025_Full - IE 343 Engineering Economics Lecture 25...

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IE 343 Engineering Economics Lecture 25: Chapter 6 - Comparison and Selection among Alternatives Instructor: Tian Ni Oct.26, 2011 IE 343 Fall 2011 1

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Case 3: useful lives > study period Solution: The strategy is to truncate the alternative at the end of the study period. We need to find the market value of the alternative at the end of the study period, and we use the imputed(implied) market value technique to do this. IE 343 Fall 2011 Case 3: useful lives > study period 2
Let T denote the study period ( T < useful life), and let MV T denote the market value at time T . Then, MV T = PW at the end of year T of remaining capital recovery(CR) amount + PW at the end of year T of original market (salvage) value at end of useful life. Formally, let I be the initial investment and S be the salvage value of a project with a useful life of U time periods where U>T . Let i %=MARR, MV T = PW T (MARR) CR + PW T (MARR) S =[ I (A/P, i %, U )- S (A/F, i %, U )](P/A, i %, U-T )+ S (P/F, i %, U-T ). IE 343 Fall 2011 Case 3: useful lives > study period 3

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Consider a project with a capital investment of \$47,600, useful life of 9 years, annual expenses of \$1,720, and market (salvage) value of \$5,000 at the end of its useful life. What is the estimated market value of this project at the end of year 5? The MARR is 20%. Note : Useful life = 9 years, Study period = 5 years, so it’s case 3: Useful life > Study Period IE 343 Fall 2011 Example 6.8 B Investment 47,600 Annual expenses 1,720 Useful life(years) 9 Salvage value 5,000 4
The estimated Imputed Market Value at the end of year 5 is: Recall the formula for computing imputed MV MV T = PW T (MARR) CR + PW T (MARR) S =[ I (A/P, i %, U )- S (A/F, i %, U )](P/A, i %, U-T )+ S (P/F, i %, U-T ) U: Useful Life = 9, T: Study Period = 5 MV 5 = [47,600(A/P,20%,9)-5,000(A/F,20%,9)](P/A,20%,4) +5,000(P/F,20%,4) = \$32,358. IE 343 Fall 2011 Example 6.8 – Solution 5

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IE 343 Fall 2011 Example 6.8 0 \$47,600 A=\$1,720 A A 1 2 3 4 5 6 7 8 9 A A ………………………. .…. . Study period = 5 years S=\$5,000 MARR = 20% 6
IE 343 Fall 2011 Example 6.8 – Solution 0 \$47,600 A=\$1,720 A A 1 2 3 4 5 A Study period = 5 years MARR = 20% Alternative B MV5 = \$32,361 7

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Which of the following alternatives should be selected if the study period is 5 years and the MARR is 20%? Alternative B is identical to the project in Example 6.8. The imputed market value of Alternative B at the end of year 5 was found to be \$32,358. IE 343 Fall 2011 Example 6.9 A B Investment 33,200 47,600 Annual expenses 2,165 1,720 Useful life(years) 5 9 Salvage value 0 5,000 8
We can use the AW method to compare the alternatives: AW(20%) A = -33,200(A/P,20%,5)-2,165 = -\$13,266. AW(20%)

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Lec%2025_Full - IE 343 Engineering Economics Lecture 25...

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